Question

Five men go to a restaurant together and each of them orders a dish that is different from the dishes ordered by the other members of the group. However, the waiter serves the dishes randomly. Then the probability that exactly one of them gets the dish he ordered equals __________ (round off to 2 decimal places)

Answer 1

Correct Answer: 0.35

To solve this problem, we calculate the probability that exactly one out of five men receives the correct dish. Let's define:
Total ways to distribute dishes: There are five unique dishes for five people, so the total number of ways to distribute them is \(5!\) (factorial of 5), which is 120.
Finding favorable outcomes: We need exactly one person to get the correct dish, while others do not. Pick 1 person out of 5, which gives us \( \binom{5}{1} = 5 \) ways.
For the remaining 4 people, none should get their own dish. This is a classic derangement problem where no element appears in its original position. The formula for the number of derangements (\(!n\)) of \(n\) items is given by:
\(!n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}\)
For \(n=4\), \(!4 = 4! \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\right) = 24 \left(1 - 1 + 0.5 - \frac{1}{6} + \frac{1}{24}\right) = 9\).
The number of favorable outcomes where exactly one man gets the right dish is \(5 \times 9 = 45\).
Calculating probability: The probability is the ratio of favorable outcomes to total outcomes, i.e., \( \frac{45}{120} = \frac{3}{8} = 0.375\).